With the decrease of current in the primary coil from 2 amperes to zero value in 0.01 s the emf generated in the secondary coil is 1000 volts. The mutual inductance of the two coils is

Options

A. 1.25 H
B. 2.50 H
C. 5.00 H
D. 10.00 H

Answer

C

Explanation

Induced emf `epsilon = (-Mdl_(P))/(dt)`
where M is the mutual inductance of two coils and `(dI_(P))/(dt)` is the rate of change of current in the primary.
`therefore 1000 V = (-M(OA-2A))/(0.01 s)`
or `M = ((1000 V)(0.01 s))/((2 A))=5 H`.